Question

The common chord of the circle $x^2+y^2+8x+4y-5=0$ and circle passing through the origin and touching the line $y=x$, passes through the fixed point

• A. $(\frac{5}{12},\frac{5}{12})$
• B. $(\frac{5}{12},-\frac{5}{12})$
• C. $(-\frac{5}{12},\frac{5}{12})$
• D. None of these

Answer 1

The correct option is A $(\frac{5}{12},\frac{5}{12})$

Given circle is $x^2+y^2+8x+4y-5=0$ ...(1)

Let the equation of the second circle be $x^2+y^2+2gx+2fy+c=0$

Since it passes through origin $\therefore c=0$

So, the equation becomes $x^2+y^2+2gx+2fy=0$ ...(2)

The equation of common chord of (1) and (2) is

$2(g-4)x+2(f-2)y+5=0$ ...(3)

Since the line $y=x$ touches the circle (2)

$\therefore x^2+x^2+2gx+2fx=0$ has equal roots

$\Rightarrow f+g=0$

$\therefore$ From (3), the equation of common chord is

$2(g-4)x+2(-g-2)y+5=0\Rightarrow (-8x-4y+5)+g(2x-2y)=0$

which passes through the point of intersection of

$8x+4y-5=0$ and $x=y$

i.e. the point is $(\frac{5}{12},\frac{5}{12})$